Torque and Rotational Equilibrium Case III:
(m1)(x1) = (m2)(x2)
Mass D:
(m1)(.33) = (200)(.45)
m1=272.727g
Actual mass: 271.02g
Mass B:
(m1)(.33) = (200)(.42)
m1 = 254.54g
Actual mass: 257.51
Mass R:
(m1)(.33) = (200)(.235)
m1 = 142.42g
Actual mass: 132.40g
Mass E:
(m1)(.33) = (200)(.37)
m1 = 224.24g
Actual mass: 206.14g
Mass F:
(m1)(.33) = (200)(.325)
m1 = 196.97
Actual mass: 196.7
Percent Error:
Mass D:
(272.727-271.02)/(271.02) x100 = .63%
Mass B:
(254.54-257.71)/(257.71) x100 = 1.23%
Mass R:
(142.42-132.40)/(132.40) x100 = 7.57%
Mass E:
(224.24-206.14)/(206.14) x100 = 8.78%
Mass F:
(196.97-196.7)/(196.7) x100 = .137%
Case IV
(m1)(x1) = (m2)(x2)
(100)(.239) = (m2)(.261)
m2 = 91.57g
Mass of meter stick is equal to 96.43g
Percent Error: .05%
Questions
1) A somewhat small trivial force can produce a large or above average torque while a large/ginormous force can produce a fairly arbitrary amount of torque. The reason why this is so is because Torque = Lever arm * Force. Henceforth, if there happens to be a generally large force, to then make it a generally small torque, the lever arm as we like to call it is decreased. For a torque to be enormous from a tiny force, the lever arm has to be most deffinitely bigger.
2) The force that is down and the force that is up acts on the one meter long stick cancelling each other out, so the balances become equal. The forces directly and not indirectly balance each other out, so the force of the one meter long stick just so happens to be irrelevant. Any change in mass, even a minute change at the fulcrum as amazing as that sounds would not change the balance of these fantastic masses.
3) (m1)(x1) = (m2)(x2) + (m3)(x3)
(220)(.35) = (m2)(100) + (.5)(120)
Mass of the meter stick = 17g
Case III:
(m1)(x1) = (m2)(x2)
Mass D:
(m1)(.33) = (200)(.45)
m1=272.727g
Actual mass: 271.02g
Mass B:
(m1)(.33) = (200)(.42)
m1 = 254.54g
Actual mass: 257.51
Mass R:
(m1)(.33) = (200)(.235)
m1 = 142.42g
Actual mass: 132.40g
Mass E:
(m1)(.33) = (200)(.37)
m1 = 224.24g
Actual mass: 206.14g
Mass F:
(m1)(.33) = (200)(.325)
m1 = 196.97
Actual mass: 196.7
Percent Error:
Mass D:
(272.727-271.02)/(271.02) x100 = .63%
Mass B:
(254.54-257.71)/(257.71) x100 = 1.23%
Mass R:
(142.42-132.40)/(132.40) x100 = 7.57%
Mass E:
(224.24-206.14)/(206.14) x100 = 8.78%
Mass F:
(196.97-196.7)/(196.7) x100 = .137%
Case IV
(m1)(x1) = (m2)(x2)
(100)(.239) = (m2)(.261)
m2 = 91.57g
Mass of meter stick is equal to 96.43g
Percent Error: .05%
Questions
1) A somewhat small trivial force can produce a large or above average torque while a large/ginormous force can produce a fairly arbitrary amount of torque. The reason why this is so is because Torque = Lever arm * Force. Henceforth, if there happens to be a generally large force, to then make it a generally small torque, the lever arm as we like to call it is decreased. For a torque to be enormous from a tiny force, the lever arm has to be most deffinitely bigger.
2) The force that is down and the force that is up acts on the one meter long stick cancelling each other out, so the balances become equal. The forces directly and not indirectly balance each other out, so the force of the one meter long stick just so happens to be irrelevant. Any change in mass, even a minute change at the fulcrum as amazing as that sounds would not change the balance of these fantastic masses.
3) (m1)(x1) = (m2)(x2) + (m3)(x3)
(220)(.35) = (m2)(100) + (.5)(120)
Mass of the meter stick = 17g